We will determine the ration of borneol to isoborneol formed by the reduction of

Question

We will determine the ration of borneol to isoborneol formed by the reduction of camphor using 1H NMR. In borneol the exo hydrogen on the ring carbon bearing the hydroxyl group appears at 4.0 ppm, whereas the analogous hydrogen in isoborneol appears at 3.6 ppm. Suppose you isolate 0.264 g of product, and integration of the key hydrogen signals yields these values: H at 4.0 ppm: 1.00 H at 3.6 ppm: 3.14 A. How many moles of borneol/isoborneol mixture were formed? B. What percentage of the mixture is borneol? What percentage is isoborneol? C. How many moles of borneol were formed? Of isoborneol?

Explanation / Answer

molar mass borneol = 154.25 g/mol

molar mass isoborneol = 154.25 g/mol

mass of product = 0.264 g

A. mass borneol in mixture = 1/4.14 = 0.064 g

moles borneol = 0.064/154.25 = 0.000415 moles

mass isoborneol in mixture = 3.14/4.14 = 0.200 g

moles isoborneol in mixture = 0.200/154.25 = 0.0013 moles

Total area of peak = 1 : 3.14 = 4.14

B. % borneol = 1 x 100/4.14 = 24.2%

% isoborneol = 3.14 x 100/4.14 = 75.8%

C.

moles borneol = 0.064/154.25 = 0.000415 moles

moles isoborneol in mixture = 0.200/154.25 = 0.0013 moles

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